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Solve the system of equations by substitution. $\newline$ $3x + 3y - 2z = 18$ $\newline$ $x + 2y + 2z = -9$ $\newline$ $z = -9$

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Solve a system of equations in three variables using substitution

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Q. Solve the system of equations by substitution.

\newline

3x + 3y - 2z = 18

\newline

x + 2y + 2z = -9

\newline

z = -9

Substitute $z = -9$ : Substitute $z = -9$ into the second equation $x + 2y + 2z = -9$ . $\newline$ $x + 2y + 2(-9) = -9$ $\newline$ $x + 2y - 18 = -9$ $\newline$ $x + 2y = 9$
Substitute $z = -9$ : Substitute $z = -9$ into the first equation $3x + 3y - 2z = 18$ . $\newline$ $3x + 3y - 2(-9) = 18$ $\newline$ $3x + 3y + 18 = 18$ $\newline$ $3x + 3y = 0$
Divide and simplify: Divide the third equation by $3$ to simplify. $\newline$ $x + y = 0$
Subtract to find $y$ : Now we have two equations: $\newline$ $1$ ) $x + 2y = 9$ $\newline$ $2$ ) $x + y = 0$ $\newline$ Subtract the second equation from the first to find $y$ . $\newline$ $(x + 2y) - (x + y) = 9 - 0$ $\newline$ $x + 2y - x - y = 9$ $\newline$ $y = 9$
Substitute $y$ to find $x$ : Substitute $y = 9$ into the second equation $x + y = 0$ to find $x$ .
$x + 9 = 0$
$x = -9$
Final values: We have found the values: $\newline$ $x = -9$ $\newline$ $y = 9$ $\newline$ $z = -9$

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